题目
Problem
The plane Π \Pi Π r = j + 2 k + s ( i − j ) + t ( − 3 i + j − k ) \mathbf{r}=\mathbf{j}+2\mathbf{k}+s(\mathbf{i}-\mathbf{j})+t(-3\mathbf{i}+\mathbf{j}-\mathbf{k}) r = j + 2 k + s ( i − j ) + t ( − 3 i + j − k )
(a) Obtain an equation of Π \Pi Π a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
(b) The point P P P ( 6 + 6 θ ) i + ( 1 + θ ) j + ( 2 + 5 θ ) k (6+6\theta)\mathbf{i}+(1+\theta)\mathbf{j}+(2+5\theta)\mathbf{k} ( 6 + 6 θ ) i + ( 1 + θ ) j + ( 2 + 5 θ ) k
Find the set of values of θ \theta θ P P P Π \Pi Π 6 \sqrt{6} 6
(4)
(c) Find an equation for the common perpendicular to the lines
r = λ ( j + 2 k ) and r = i + 2 k + μ ( − 3 i + j − k ) . \begin{align*}
\mathbf{r}=&\,\lambda(\mathbf{j}+2\mathbf{k})\quad\text{and}\\
\mathbf{r}=&\,\mathbf{i}+2\mathbf{k}+\mu(-3\mathbf{i}+\mathbf{j}-\mathbf{k}).
\end{align*} r = r = λ ( j + 2 k ) and i + 2 k + μ ( − 3 i + j − k ) . (8)
题目中文翻译
平面 Π \Pi Π r = j + 2 k + s ( i − j ) + t ( − 3 i + j − k ) \mathbf{r}=\mathbf{j}+2\mathbf{k}+s(\mathbf{i}-\mathbf{j})+t(-3\mathbf{i}+\mathbf{j}-\mathbf{k}) r = j + 2 k + s ( i − j ) + t ( − 3 i + j − k )
(a) 求 Π \Pi Π a x + b y + c z = d ax+by+cz=d a x + b y + cz = d
(4)
(b) 点 P P P ( 6 + 6 θ ) i + ( 1 + θ ) j + ( 2 + 5 θ ) k (6+6\theta)\mathbf{i}+(1+\theta)\mathbf{j}+(2+5\theta)\mathbf{k} ( 6 + 6 θ ) i + ( 1 + θ ) j + ( 2 + 5 θ ) k
求使 P P P Π \Pi Π 6 \sqrt{6} 6 θ \theta θ
(4)
(c) 求下列两条直线的公垂线方程：
r = λ ( j + 2 k ) , r = i + 2 k + μ ( − 3 i + j − k ) . \begin{align*}
\mathbf{r}=&\,\lambda(\mathbf{j}+2\mathbf{k}),\\
\mathbf{r}=&\,\mathbf{i}+2\mathbf{k}+\mu(-3\mathbf{i}+\mathbf{j}-\mathbf{k}).
\end{align*} r = r = λ ( j + 2 k ) , i + 2 k + μ ( − 3 i + j − k ) . (8)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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用两条平面内方向的叉积求法向量，再代入平面上给定点确定常数。
答题过程
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Two direction vectors in Π \Pi Π u = ( 1 , − 1 , 0 ) \mathbf u=(1,-1,0) u = ( 1 , − 1 , 0 ) v = ( − 3 , 1 , − 1 ) \mathbf v=(-3,1,-1) v = ( − 3 , 1 , − 1 )
n = u × v = ( 1 , 1 , − 2 ) . \mathbf n=\mathbf u\times\mathbf v=(1,1,-2). n = u × v = ( 1 , 1 , − 2 ) . The plane contains ( 0 , 1 , 2 ) (0,1,2) ( 0 , 1 , 2 )
x + y − 2 z = 0 + 1 − 2 ( 2 ) , x + y − 2 z = − 3 . x+y-2z=0+1-2(2),
\qquad\boxed{x+y-2z=-3}. x + y − 2 z = 0 + 1 − 2 ( 2 ) , x + y − 2 z = − 3 .
(b)
解法一
思路
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代入点到平面的距离公式，再解严格绝对值不等式；平面方程常数项的符号要特别注意。
答题过程
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The perpendicular distance is
d = ∣ 6 + 6 θ + 1 + θ − 4 − 10 θ + 3 ∣ 6 = ∣ 6 − 3 θ ∣ 6 . \begin{align*}
d=&\,\frac{|6+6\theta+1+\theta-4-10\theta+3|}{\sqrt6} \\[4mm]
=&\,\frac{|6-3\theta|}{\sqrt6}.
\end{align*} d = = 6 ∣6 + 6 θ + 1 + θ − 4 − 10 θ + 3∣ 6 ∣6 − 3 θ ∣ . Hence d < 6 d<\sqrt6 d < 6
∣ 6 − 3 θ ∣ < 6 , − 6 < 6 − 3 θ < 6. |6-3\theta|<6,
\qquad -6<6-3\theta<6. ∣6 − 3 θ ∣ < 6 , − 6 < 6 − 3 θ < 6. Solving gives 0 < θ < 4 \boxed{0<\theta<4} 0 < θ < 4
(c)
解法一
思路
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设两条线上的点Q、R，公垂线连接向量同时垂直于两方向。解参数后用最近点与公垂线方向写直线方程。
答题过程
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Let
Q = ( 0 , λ , 2 λ ) , R = ( 1 − 3 μ , μ , 2 − μ ) \begin{align*}
Q=&\,(0,\lambda,2\lambda),\\[4mm]
R=&\,(1-3\mu,\mu,2-\mu)
\end{align*} Q = R = ( 0 , λ , 2 λ ) , ( 1 − 3 μ , μ , 2 − μ ) lie on the first and second lines, respectively. Then
Q R → = ( 1 − 3 μ , μ − λ , 2 − μ − 2 λ ) . \overrightarrow{QR}=(1-3\mu,\mu-\lambda,2-\mu-2\lambda). QR = ( 1 − 3 μ , μ − λ , 2 − μ − 2 λ ) . For Q R QR QR
Q R → ⋅ ( 0 , 1 , 2 ) = 0 , Q R → ⋅ ( − 3 , 1 , − 1 ) = 0. \begin{align*}
\overrightarrow{QR}\cdot(0,1,2)=&\,0, \\[4mm]
\overrightarrow{QR}\cdot(-3,1,-1)=&\,0.
\end{align*} QR ⋅ ( 0 , 1 , 2 ) = QR ⋅ ( − 3 , 1 , − 1 ) = 0 , 0. These give
μ + 5 λ = 4 , 11 μ + λ = 5. \begin{align*}
\mu+5\lambda=&\,4,\\[4mm]
11\mu+\lambda=&\,5.
\end{align*} μ + 5 λ = 11 μ + λ = 4 , 5. From λ = 5 − 11 μ \lambda=5-11\mu λ = 5 − 11 μ μ + 25 − 55 μ = 4 \mu+25-55\mu=4 μ + 25 − 55 μ = 4
μ = 7 18 , λ = 13 18 . \mu=\frac7{18},\qquad\lambda=\frac{13}{18}. μ = 18 7 , λ = 18 13 . Therefore
R = 1 18 ( − 3 , 7 , 29 ) , Q R → = − 1 6 ( 1 , 2 , − 1 ) . R=\frac1{18}(-3,7,29),
\qquad\overrightarrow{QR}=-\frac16(1,2,-1). R = 18 1 ( − 3 , 7 , 29 ) , QR = − 6 1 ( 1 , 2 , − 1 ) . The common perpendicular is
r = 1 18 ( − 3 i + 7 j + 29 k ) + t ( i + 2 j − k ) , t ∈ R . \begin{align*}
\mathbf r=&\,\frac1{18}(-3\mathbf i+7\mathbf j+29\mathbf k)\\[4mm]
&\,+t(\mathbf i+2\mathbf j-\mathbf k),\qquad t\in\mathbb R.
\end{align*} r = 18 1 ( − 3 i + 7 j + 29 k ) + t ( i + 2 j − k ) , t ∈ R . Its direction has zero dot product with ( 0 , 1 , 2 ) (0,1,2) ( 0 , 1 , 2 ) ( − 3 , 1 , − 1 ) (-3,1,-1) ( − 3 , 1 , − 1 )
